two resistors of 25 ohms and 5 ohms respectively are connected in series to a 240V supply calculate the value of a third resistor to be connected in parallel with 25 ohms resistor so that the power dissipated shall be tripled

Answer 1

For each branch, P = E^2/R

Currently (no pun intended), we are dissipating 240^2/(25+5) = 1920 watts

So, we want to dissipate 3840 watts in the other branch.

240^2/R = 3840
R = 15 ohms

Answer 2

two resistors in series, 25,5 ohms.
power= 1920
want 240^2/(r'+5)=3840
figure r', then

1/r'=1/25+1/R
solve for R, the resistance going in parallel with the 25 ohm resistor

Answer 3

Go with bobpursley -- I misread the problem, connecting the parallel R across both series resistances.

Answer 4

P = E^2/(R1+R2) = 240^2/(25+5) = 1920 Watts = Current power.

Rt = (25+5)/3 = 10 Ohms = Total resistance required to triple the power.

Rt = R1*R3/(R1+R3) + R2 = 10.
(25*R3)/(25+R3) + 5 = 10
25R3/(25+R3) = 5
25R3 = 125 + 5R3
R3 = 6.25 Ohms.